Solution:

Let's start by finding the prime factorization of 30 which is $2 \cdot 3 \cdot 5$. There are 4 factors of 30 that are not composite, namely 1, 2, 3, and 5. We can use these factors to construct other numbers that have 4 non-composite factors.

Step 1: Finding the number of factors

Let the number be represented by $n = p_1^{a_1} \cdot p_2^{a_2} \cdot ... \cdot p_k^{a_k}$, where $p_1, p_2, ..., p_k$ are distinct primes and $a_1, a_2, ..., a_k$ are positive integers. The number of factors of $n$ is given by $(a_1 + 1)(a_2 + 1) \cdot ... \cdot (a_k + 1)$. Therefore, we need to find the prime factorization of $n$ such that $(a_1 + 1)(a_2 + 1) \cdot ... \cdot (a_k + 1) = 32$.

Step 2: Finding the number of non-composite factors

If $n$ has 32 factors, then the product of the non-composite factors must be a factor of $n$. Since the product of the non-composite factors is 30, we know that $n$ must be divisible by 30.

Step 3: Finding the prime factorization of $n$

Let $n = 30m$, where $m$ is a positive integer. Then, $(a_1 + 1)(a_2 + 1) \cdot ... \cdot (a_k + 1) = (2 + 1)(1 + 1)(1 + 1)(a_4 + 1) \cdot ... \cdot (a_k + 1) = 32/3$. Since each factor in the product is at least 2, we know that $k \geq 5$. Therefore, $n$ must have at least 5 distinct prime factors.

Step 4: Counting the number of possible values of $n$

Let $n = p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3} \cdot p_4^{a_4} \cdot p_5^{a_5}$, where $p_1, p_2, p_3, p_4, p_5$ are distinct primes and $a_1, a_2, a_3, a_4, a_5$ are positive integers. Since $n$ must have at least 5 distinct prime factors, there are 3 cases to consider:

Case 1: $n$ has exactly 5 distinct prime factors

In this case, we have $(a_1 + 1)(a_2 + 1)(a_3 + 1)(a_4 + 1)(a_5 + 1) = 32/3$. Since each factor in the product is at least 2, we know that $a_5 \geq 2$. Therefore, we can